iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :
A.
C6H8N
B.
C6H8N2
C.
C12H8N
D.
C12H8N2
Correct Answer: B
Explanation:
Molar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1
Thus, the correct formula is C6H8N2.
.
2019
Q254
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol–1]
A.
33.6
B.
0.84
C.
8.4
D.
16.8
Correct Answer: C
Explanation:
2NaHCO3
+ (COOH)2 $ \to $ (COONa)2 + 2H2O + 2CO2
$ \therefore $ 1 mole of CO2 is produced by 1 mole of NaHCO3.
Given, volume of CO2 produced = 0.25 ml
25 L of CO2 contains 1 mol
$ \therefore $ 0.25 ml of CO2 contains = ${1 \over {25 \times {{10}^3}}} \times 0.25$ moles
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg$-$1 is :
A.
12
B.
4
C.
8
D.
16
Correct Answer: B
Explanation:
Molality of Na+ = ${{Moles\,\,of\,\,N{a^ + }} \over {mass\,\,of\,\,{H_2}O\,\,in\,\,kg}}$
Moles of Na+ = ${{92} \over {23}}$ = 4
Mass of H2O = 1 kg
$ \therefore $ Molality = ${4 \over 1}$ = 4
2019
Q258
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm$-$3, then molarity of urea solution is ................
(Given data : Molar masses of urea and water are 60 g mol$-$1 and 18 g mol$-$1, respectively)
$ = {{2.6315 \times 1000} \over {881.57}}$ = 2.98 M
2019
Q259
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO3 to a compound with the highest oxidation state of sulphur is ..............
(Given data : Molar mass of water = 18 g mol$-$1)
Correct Answer: 288
Explanation:
When rhombic sulphur (S8) is oxidised by conc. HNO3 then H2SO4 is obtained and NO2 gas is released.
1 mole of rhombic sulphur produces = 16 moles of H2O
$ \therefore $ Mass of water = 16 $ \times $ 18 (molar mass of H2O) = 288 g
2018
Q260
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $ \times $ 1023 mol-1)
A.
6.023 $ \times $ 1020
B.
6.023 $ \times $ 109
C.
6.023 $ \times $ 1021
D.
6.023 $ \times $ 1023
Correct Answer: A
Explanation:
% of Cl = 3.55
$\therefore\,\,\,\,$ In 100 g chlorohydrocarbon 3.55 gm Cl present.
In 1 gm chlorohydrocarbon Cl present
= ${{3.55} \over {100}}$
= 0.0355 gm
$\therefore\,\,\,\,$ No of Moles of Cl = ${{0.0355} \over {35.5}}$
= 0.001 mole
$\therefore\,\,\,\,$ no of Cl atoms = 0.001 $ \times $ 6.023 $ \times $ 103
= 6.023 $ \times $ 1020
2018
Q261
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is
A.
C2H4O3
B.
C3H6O3
C.
C2H4O
D.
C3H4O2
Correct Answer: A
Explanation:
$\therefore\,\,\,$ The ratio of no of atoms of C and H in one molecule of Cx Hy Oz = 1 : 2
$\therefore\,\,\,$ y = 2x
In one molecule of Cx Hy Oz compound contain z atoms of oxygen.
According to the question, no of atoms of oxygen required to burn CxHy completely should be twice of z atoms of oxygen.
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
For per gram of reactant, the maximum quantity of N2 gas is produced in which of the following thermal decomposition reactions ?
(Given : Atomic wt. - Cr = 52 u, Ba = 137 u)
A.
(NH4)2Cr2O7(s) $ \to $ N2(g) + 4H2O(g) + Cr2O3(s)
B.
2NH4NO3(s) $ \to $ 2 N2(g) + 4H2O(g) + O2(g)
C.
Ba(N3)2(s) $ \to $ Ba(s) + 3N2(g)
D.
2NH3(g) $ \to $ N2(g) + 3H2(g)
Correct Answer: D
Explanation:
(a) Molar mass of (NH4)2 Cr2O7 = 252 g/mol.
252g of (NH4)2 Cr2 O7 produce 28g mole of N2
$\therefore\,\,\,$ 1 g of (NH4)2 Cr2 O7 Produce = ${{28} \over {252}} = 0.111$ g N2
(b) Molar mass of NH4 NO3 = 80 g/mol
2 $ \times $ 80 g of NH4 NO3 produce 28 $ \times $ 2g of N2
$\therefore\,\,\,$ 1 g of NH4 NO3 produce = ${{28 \times 2} \over {2 \times 80}}$ = 0.35 g N2
(c) Molar mass of Ba(N3)2 = 221 g/mol
221 g of Ba(N3)2 produce 3 $ \times $ 28 g of N2
$\therefore\,\,\,$ 1 g of Ba(N3)2 produce = ${{3 \times 28} \over {221}}$ = 0.38 g of N2
(d) Molar mass of NH3 = 17 g/mol
17 $ \times $ 2 NH3 produce 28 g of N2
$\therefore\,\,\,$ 1 g of NH3 produce = ${{28} \over {17 \times 2}}$ = 0.823 g of N2
2018
Q263
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A sample of $NaCl{O_3}$ is converted by heat to $NaCl$ with a loss of $0.16$ $g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl.$ The mass of $AgCl$ (in $g$) obtained will be : (Given : Molar mass of $AgCl=143.5$ $g$ $mo{l^{ - 1}}$)
Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid ($225$ $mg$) was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnC{l_2}$ (in mg) present in the initial solution is ____________.
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must
be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid
solution ?
A.
320
B.
325
C.
316
D.
330
Correct Answer: C
Explanation:
Let the volume of a monoprotic acid solution in mL be V
1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3
$\therefore\,\,\,$ 2 $ \times $ 10$-$2 moles of Fe(OH)3 will obtain from
= 0.02 mole of FeCl3
Molarity of FeCl3 = ${{No.of\,moles} \over {Volume\,in\,L}}$ = ${{2 \times {{10}^{ - 2}}} \over {0.1}}$ = 0.2 M
2017
Q267
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon
(22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all
1H atoms are replaced by 2H atoms is:
A.
37.5 kg
B.
7.5 kg
C.
10 kg
D.
15 kg
Correct Answer: B
Explanation:
Given that weight of human adult is = 75 kg
Among those 75 kg, 10% is Hydrogen(1H).
$\therefore$ Mass of 1H = $75 \times {{10} \over {100}}$ = 7.5 kg
Now when every 1H atom is replaced by 2H atom then weight of every atom is become double. So total weight of 2H becomes = 2$ \times $7.5 = 15 kg.
So the weight gain by the person = 15 - 7.5 = 7.5 kg
2017
Q268
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is:
For this reaction, 2KClO3 $\buildrel \, \over
\longrightarrow $ 2KCl + 3O2
We can write, ${{{n_{KCl{O_3}}}} \over 2} = {{{n_{KCl}}} \over 2} = {{{n_{{O_2}}}} \over 3}$
This means
2016
Q269
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
An organic compound contains C, H and S. The minimum molecular weight of the
compound containing 8% sulphur is :
(atomic weight of S = 32 amu)
A.
200 g mol$-$1
B.
400 g mol$-$1
C.
600 g mol−1
D.
300 g mol−1
Correct Answer: B
Explanation:
We know that 8% sulphur means 8 g of sulphur present in 100 g of an organic compound.
Thus, 32 g of sulphur present in ${{100} \over 8} \times 32g$ = 400 g of organic compound.
Therefore, minimum molecular weight of the compound is 400 g mol$-$1.
2016
Q270
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The molecular formula of a commercial resin used for exchanging ions in water softening is
C8H7SO3Na (Mol. Wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?
A.
2/309
B.
1/412
C.
1/103
D.
1/206
Correct Answer: B
Explanation:
$2$ mole of water softner require $1$ mole of $C{a^{2 + }}\,\,$ ion
So, $1$ mole of water softner require ${1 \over 2}$ mole of $C{a^{2 + }}\,$ ion
Thus, ${1 \over {2 \times 206}} = {1 \over {412}}mol/g\,\,\,$ will be maximum uptake
2014
Q272
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of
their molecule is:
A.
1: 8
B.
3 : 16
C.
1 : 4
D.
7 : 32
Correct Answer: D
Explanation:
Given ratio of mass of O2 and N2 = 1 : 4
Let mass of O2 = w
and mass of N2 = 4w
$\therefore$ Number of moles of O2 = ${w \over {32}}$
Number of molecules of O2 = ${w \over {32}}$$ \times $NA
Number of moles of N2 = ${4w \over {28}}$
Number of molecules of N2 = ${4w \over {28}}$$ \times $NA
$\therefore$ Ratio of molecules of O2 and N2 = ${w \over {32}}$$ \times $NA : ${4w \over {28}}$$ \times $NA
= 7 : 32
2014
Q273
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the value of Avogadro number is 6.023 $\times$ 1023 mol-1 and the value of Boltzmann constant is 1.380 $\times$ 10-23 J K-1, then the number of significant digits in the calculated value of the universal gas constant is
Correct Answer: 4
Explanation:
The universal gas constant, denoted by R, can be calculated using the Avogadro number (NA) and the Boltzmann constant (kB) by the following relationship:
To determine the number of significant digits in the calculated value of R, we must consider the number of significant digits in the given values of NA and kB.
The value for NA has four significant digits (6.023), and the value for kB also has four significant digits (1.380). When multiplying or dividing numbers, the number of significant digits in the result is determined by the number with the smallest amount of significant digits used in the calculation.
In this case, since both constants have four significant digits, the value of R calculated from their multiplication will also contain four significant digits:
$
R \approx 8.314 \text{ J mol}^{-1} \text{K}^{-1}
$
Therefore, the calculated value of the universal gas constant R has four significant digits.
2014
Q274
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A compound H2X with molar weight of 80g is dissolved in a solvent having density of 0.4 gml–1 . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is
Correct Answer: 8
Explanation:
Understanding the Concepts
Molarity (M): Moles of solute per liter of solution.
Molality (m): Moles of solute per kilogram of solvent.
Density: Mass per unit volume.
Steps to Calculate Molality
Find the Mass of Solute (H₂X):
Molarity = moles of solute / volume of solution (in liters)
3.2 M solution means 3.2 moles of H₂X are present in 1 liter of solution.
Mass of H₂X = moles $ \times $ molar mass = 3.2 moles $ \times $ 80 g/mole = 256 g
Find the Mass of Solvent:
Assuming no change in volume, the volume of the solution remains 1 liter.
Density = mass / volume
Mass of solvent = density $ \times $ volume = 0.4 g/mL $ \times $ 1000 mL = 400 g
Convert Mass of Solvent to Kilograms:
1 kg = 1000 g
Mass of solvent = 400 g $ \times $ (1 kg / 1000 g) = 0.4 kg
Calculate Molality:
Molality = moles of solute / mass of solvent (in kg)
Molality = 3.2 moles / 0.4 kg = 8 mol/kg
Answer:
The molality of the 3.2 molar solution is 8 mol/kg.
2013
Q275
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is
0.72 gm of H2O = ${{{0.72} \over {18}}}$ mole of H2O = 0.04 mole of H2O
In one H2O molecule 2 hydrogen atoms present.
So in 0.04 mole of H2O molecules 2$ \times $0.04 = 0.08 moles of H atoms present.
0.72 gm of CO2 = ${{{3.08} \over {44}}}$ mole of CO2 = 0.07 mole of CO2
And in one CO2 molecule 1 C atom present.
So in 0.07 mole of CO2 molecules 0.07$\times$1 = 0.07 moles of C atoms present
$\therefore$ C : H = 0.07 : 0.08 = 7 : 8
$\therefore$ Empirical formula of hydrocarbon = C7H8
2013
Q276
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be
A.
7.01%
B.
4.08%
C.
6.05%
D.
5.08%
Correct Answer: B
Explanation:
Given metal oxide = M0.98O
We know oxidation number of O = (-2)
Now assume oxidaton no of M = $x$
$\therefore$ 0.98$x$ + 1$ \times $ (-2) = 0
$ \Rightarrow x =$ ${{200} \over {98}}$
This represent the charge in one atom of M. As you can see the charge of M is in the range $2 < {{200} \over {98}} < 3$.
So we can say in M mixture of M+2 and M+3 present.
Assume total no of atoms present in M is 100.
Let M+3 present in M = y atoms
So M+2 present in M = (100 - y) atoms
In 1 atom of M+3 charge present = +3
So in y atoms of M+3 charge present = +3y
Similarly in 1 atom of M+2 charge present = +2
So in (100 - y) atoms of M+2 charge present = +2(100 - y)
$\therefore$ Total charge = 200 - 2y + 3y = 200 + y
In 100 atoms of M total charge = 200 + y
So in 1 atoms of M total charge = ${{200 + y} \over {100}}$
Earlier we found that charge in one atom of M is = ${{200} \over {98}}$
So we can write,
${{200 + y} \over {100}}$ = ${{200} \over {98}}$
$ \Rightarrow y = 4.08$
So Option (B) is correct.
2013
Q277
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:
A.
1.00 M
B.
1. 75 M
C.
0.975 M
D.
0.875 M
Correct Answer: D
Explanation:
The formula for molarity of mixture of two substance is = ${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is
1.15 g/mL. The molarity of this solution is :
A.
0.50 M
B.
1.78 M
C.
1.02 M
D.
2.05 M
Correct Answer: D
Explanation:
We know molarity (M) = ${{no\,of\,moles\,of\,solute} \over {volume\,of\,solution\,in\,litre}}$
Moles of solute = ${{120} \over {60}}$ = 2
Mass of solution = 1000 + 120 = 1120 gm
Density of solution = 1.15 g/mL
$\therefore$ volume of solution = ${{1120} \over {1.15}}$ = 973.9 mL = ${{973.9} \over {1000}}$ litre = 0.9739 litre
$\therefore$ M = ${2 \over {0.9739}}$ = 2.05
2012
Q279
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1 . The molecular weight of HCl is 36.5 g mol-1 .
The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is
Correct Answer: 8
Explanation:
To calculate the volume of the 29.2% (w/w) HCl stock solution needed to prepare a 200 mL solution of 0.4 M HCl, we need to use several steps involving concentration and density conversions.
First, we calculate the mass of HCl that is contained in the 200 mL of a 0.4 M solution:
$ Mass = Molarity \times Volume \times Molecular\ Weight $
$ Mass = 0.4 \ mol/L \times 0.200 \ L \times 36.5 \ g/mol $
Note that we convert the volume from mL to L to match the units of molarity (mol/L).
Now, we calculate it:
$ Mass = (0.4 \times 0.200 \times 36.5) \ g $
$ Mass = 0.08 \times 36.5 \ g $
$ Mass = 2.92 \ g $
The next step is to determine how much of the stock solution is needed to get 2.92 g of HCl. Since the stock solution is 29.2% (w/w) HCl, this means that in every 100 g of stock solution, there is 29.2 g of HCl. We can set up a proportion to find the mass of the stock solution needed:
$ \frac{29.2\ g \ HCl}{100\ g \ stock\ solution} = \frac{2.92\ g \ HCl}{x\ g \ stock\ solution} $
Now we solve for $ x $:
$ x = \frac{2.92\ g \times 100\ g \ stock\ solution}{29.2\ g \ HCl} $
$ x = \frac{292}{29.2} \ g $
$ x = 10\ g $
So, we need 10 g of the stock solution to get 2.92 g of HCl.
The final step is to calculate the volume of the stock solution that has a mass of 10 g. We use the density to convert mass to volume:
$ Volume = \frac{Mass}{Density} $
The density of the stock solution is given as 1.25 g/mL, so:
$ Volume = \frac{10\ g}{1.25\ g/mL} $
$ Volume = 8\ mL $
Therefore, to prepare a 200 mL solution of 0.4 M HCl, you would need to measure out 8 mL of the 29.2% HCl stock solution.
2011
Q280
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl
alcohol in the solution ?
A.
0.190
B.
0.086
C.
0.050
D.
0.100
Correct Answer: B
Explanation:
The formula between Mole fraction and Molality is
$$m = {{{X_{solute}}} \over {{X_{solvent}}}} \times {{1000} \over {{M_{solvent}}}}$$
$m$ = molality, Xsolute = Mole fraction of solute, Xsolvent = Mole fraction of solvent, Msolvent = Molar mass of solvent
Here solute is methyl alcohol(CH3OH) and solvent is water(H2O). Here water is solvent because question says solution is aqueous.
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of $[Cr{({H_2}O)_5}Cl]C{l_2}$, as silver chloride is close to ____________.
The density of the solution $ \rho $ is given as 1.15 g/mL. Therefore, the volume in milliliters (which is equivalent to cubic centimeters) is:
$ V = \frac{1120 \text{ g}}{1.15 \text{ g/mL}} $
$ V = 973.91 \text{ mL} $
To convert milliliters to liters (since molarity is defined in terms of liters), we divide by 1000:
$ V = 0.97391 \text{ L} $
Step 3: Calculate the molarity.
Molarity ($ M $) is defined as the number of moles of solute divided by the volume of solution in liters:
$ M = \frac{n_{solute}}{V_{solution}} $
$ M = \frac{2 \text{ moles}}{0.97391 \text{ L}} $
$ M \approx 2.05 \text{ M} $
Therefore, the molarity of the urea solution is approximately 2.05 M, which corresponds to Option C.
2010
Q283
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is
Correct Answer: 3
Explanation:
To find the average titre value, first add up the three measurements provided and then divide by the number of measurements.
When reporting the average, we must consider the significant figures of the original measurements. The number of significant figures is determined by the least precise measurement, which in this case is 25.0 mL with three significant figures. Therefore, we should report the average value to three significant figures as well.
The average value of 25.15 mL has four significant figures, so we need to round it to three significant figures. However, this is slightly tricky since 25.15 already appears to be rounded to four significant figures. We should consult the original measurements to decide on the best course of action.
Looking at the individual measurements (25.2, 25.25, and 25.0), we should consider the lowest decimal place which they all have in common, which is the first decimal place. The third measurement has no second decimal place, indicating its level of precision. Thus, the number of significant figures for the average titre value should be in line with this level of precision. Since the average calculated is 25.15, when we adjust to the first decimal place for consistent significant figures, the average is 25.1 mL with three significant figures.
$
\text{Corrected Average titre value} = 25.1 \, \text{mL}
$
Therefore, the number of significant figures in the average titre value is three: 25.1 mL.
2010
Q284
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Silver (atomic weight = 108 g mol-1) has a density of 10.5 g.cm-3. The number of silver atoms on a surface of area 10-12 m2 can be expressed in scientific notation as y $\times$ 10x. The value of x is?
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is :
A.
55.85
B.
55.95
C.
55.75
D.
56.05
Correct Answer: B
Explanation:
To calculate the atomic mass of an element based on its isotopic abundances, we use a weighted average. The atomic mass reported on the periodic table is a reflection of the weighted averages of all the naturally occuring isotopes of that element. The formula to calculate the average atomic mass $ M $ of an element when given the abundance and atomic mass of each isotope is:
$M = \sum (f_i \times m_i)$
where:
$ f_i $ is the fractional percent abundance (as a decimal) of the $ i $-th isotope.
$ m_i $ is the mass number of the $ i $-th isotope.
For iron (Fe), we have three isotopes $ ^{54}Fe, ^{56}Fe, ^{57}Fe $ with abundances 5%, 90%, and 5%, respectively. To convert these percentages into decimals, we divide each by 100.
11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed
B.
6 L HCl(aq) is consumed for every 3L H2(g) produced
C.
33.6 L H2(g) is produced regardless of temperature and pressure for every mole that reacts
D.
67.2 H2(g) at STP is produced for every mole Al that reacts
Correct Answer: A
Explanation:
Option A : 6 moles of HCl(aq) produces = 3 moles of H2(g)
$\therefore$ 1 mole of HCl(aq) produces = ${3 \over 6}$ moles of H2(g) = ${1 \over 2}$ moles of H2(g)
From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.
$\therefore$ ${1 \over 2}$ moles of H2(g) occupies = ${1 \over 2} \times 22.4$ = 11.2 L
$\therefore$ Option (A) is correct.
Option B :
According to avogadro Hypothesis, volume is directly proportional to the no of moles of gases.
Only for gases, no of mole and volume relationship is possible i.e, for two gases if no of moles are same then their volume is also same. But here HCL is in aqueous form and H2 is in gaseous form so we can't find volume of HCL(aq) using no of moles of HCL(aq).
$\therefore$ Option (B) is wrong.
Option C :
2 moles of Al(s) produces = 3 moles of H2(g)
$\therefore$ 1 mole of Al(s) produces = ${3 \over 2}$ moles of H2(g)
From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.
$\therefore$ ${3 \over 2}$ moles of H2(g) occupies = ${3 \over 2} \times 22.4$ = 33.6 L
6 moles of HCl(aq) produces = 3 moles of H2(g)
$\therefore$ 1 mole of HCl(aq) produces = ${3 \over 6}$ moles of H2(g) = ${1 \over 2}$ moles of H2(g)
$\therefore$ ${1 \over 2}$ moles of H2(g) occupies = ${1 \over 2} \times 22.4$ = 11.2 L
So from every mole of HCl(aq) that reacts only 11.2 L H2(g) produces.
Form every mole of Al(s) produces 33.6 L H2(g) but from every mole of HCl(aq) that reacts only 11.2 L H2(g) produces not 33.6 L.
$\therefore$ Option (C) is wrong.
Option D :
2 moles of Al(s) produces = 3 moles of H2(g)
$\therefore$ 1 mole of Al(s) produces = ${3 \over 2}$ moles of H2(g)
From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.
$\therefore$ ${3 \over 2}$ moles of H2(g) occupies = ${3 \over 2} \times 22.4$ = 33.6 L
$\therefore$ Option (D) is wrong.
2007
Q288
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is
A.
3
B.
4
C.
5
D.
6
Correct Answer: D
Explanation:
The formula of Mohr's salt is $\mathrm{FeSO_4(NH_4)_2SO_4}$. It is a mixture of ferrous sulfate and ammonium sulfate. Ferrous sulfate is oxidized by potassium dichromate to ferric sulphate.
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will
A.
be a function of the molecular mass of the substance
as 6 is half of 12 then R.A.M will decrease twice.
2005
Q292
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture?
A.
2.70 M
B.
1.344 M
C.
1.50 M
D.
1.20 M
Correct Answer: B
Explanation:
Short Cut Method : Molarity, Normality, Average Atomic Mass, Average Molar Mass, Average Density, Average Vapour Pressure all of those values will be between the solutions which are going to be mixed.
Here molarity of first solution is 1.2 and molarity of second solution is 1.5.
So the molarity of the mixture will always be more than 1.2 and less than 1.5. Remember the molarity of mixture can't be either 1.2 or 1.5.
So from the option you an see only 1.344 M can be the right answer.
Normal method : The formula for molarity of mixture of two substance is = ${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is
A.
urea
B.
benzamide
C.
acetamide
D.
thiourea
Correct Answer: A
Explanation:
Initially total H2SO4 present = 100 mL of 0.1 M = ${{{100} \over {1000}} \times 0.1}$ mole = 0.01 mole
2NaOH + H2SO4 $ \to $ Na2SO4 + 2H2O
Let in this reaction H2SO4 required n mole
So now remaining H2SO4 = 0.01 - 0.005 = 0.005 mole . Those remaining H2SO4 will react with NH3.
2NH3 + H2SO4 $ \to $ (NH4)2SO4 Let no moles of NH3 produce through this reaction is = $x$
$\therefore$ ${x \over 2}$ = ${{0.005} \over 1}$
$ \Rightarrow x$ = 0.01
In NH3 no of N atom is 1 and H atom is 3. So no of moles of N atom in NH3 = 0.01$ \times $1 = 0.01 mole
This 0.01 mole or 0.01$ \times $14 gm N is produced from 0.3 gm unknown organic compound.
$\therefore$ % of N in unknown compound is
= ${{0.01 \times 14} \over {0.3}} \times 100$
= 46.6
% of N in urea [(NH4)2CO] = ${{14 \times 2} \over {60}} \times 100$ = 46.6 %
[ Mol weight of urea = 60]
% of N in benzamide [C6H5CONH2] = ${14 \over {121}} \times 100$ = 11.5 %
[ Mol weight of benzamide [C6H5CONH2] = 121]
% of N in acetamide [CH3CONH2] = ${14 \over {59}} \times 100$ = 23.4 %
[ Mol weight of acetamide [CH3CONH2] = 59]
% of N in thiourea [NH2CONH2] = ${{14 \times 2} \over {76}} \times 100$ = 36.8 %
[ Mol weight of thiourea [NH2CSNH2] = 76]
$\therefore$ compound is urea.
2004
Q294
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
6.02 $\times$ 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is (Avogadro constant, NA = 6.02 $\times$ 1023 mol-1)
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?
A.
67.2 L
B.
44.8 L
C.
22.4 L
D.
89.6 L
Correct Answer: A
Explanation:
The reaction is
2BCl3 + 3H2 $ \to $ 2B + 6HCL
$\therefore$ 3 moles of hydrogen produces 2 moles of boron
So 2$ \times $10.8 gm = 21.6 gm of boron is produces from 3$ \times $22.4 = 67.2 L hydrogen
2003
Q297
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution.
1 litre of mixture X + excess AgNO3 $ \to $ Y.
1 litre of mixture X + excess BaCl2 $ \to $ Z
No. of moles of Y and Z are
A.
0.01, 0.01
B.
0.02, 0.01
C.
0.01, 0.02
D.
0.02, 0.02
Correct Answer: A
2003
Q299
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Which has maximum number of atoms?
A.
24 g of C(12)
B.
56g of Fe(56)
C.
27g of Al(27)
D.
108g of Ag(108)
Correct Answer: A
Explanation:
To determine which option has the maximum number of atoms, we will use Avogadro's number and the concept of moles. Avogadro's number ($6.02 \times 10^{23}$ atoms/mol) represents the number of atoms in one mole of any substance. The number of moles of a substance is calculated by dividing the given mass of the substance by its molar mass (which is given in grams per mole and is numerically equal to the element's atomic mass for pure elements).
Let's calculate the number of moles for each option:
Option A: 24 g of C (Carbon, atomic mass = 12 g/mol)
The number of moles of C = $\frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles}$
Therefore, the number of atoms in 24 g of C = $2 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mol} = 1.204 \times 10^{24}$ atoms
Option B: 56 g of Fe (Iron, atomic mass = 56 g/mol)
The number of moles of Fe = $\frac{56 \text{ g}}{56 \text{ g/mol}} = 1 \text{ mole}$
Therefore, the number of atoms in 56 g of Fe = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms
Option C: 27 g of Al (Aluminum, atomic mass = 27 g/mol)
The number of moles of Al = $\frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole}$
Therefore, the number of atoms in 27 g of Al = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms
Option D: 108 g of Ag (Silver, atomic mass = 108 g/mol)
The number of moles of Ag = $\frac{108 \text{ g}}{108 \text{ g/mol}} = 1 \text{ mole}$
Therefore, the number of atoms in 108 g of Ag = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms
Comparing the number of atoms in each option, Option A with 1.204 $\times 10^{24}$ atoms has the maximum number of atoms.
2003
Q300
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Calculate the molarity of water if it's density is 1000 kg/m3
Correct Answer: 55.55 M
Explanation:
Molarity is
calculated by dividing the number of moles of solute in the solution by the volume of the solution in liters.
Given, 1 L of water = 1 kg = 1000 g (because density = 1000
kg m−3).
